Question

Suppose a laboratory has a 31 g sample of polonium-210. The half-life of polonium-210 is about 138 days. How many half-lives of polonium-210 occur in 966 days? How much polonium is in the sample 966 days later?

Answer 1

A=P(1/2)^(t/h)
P=initial amount
t=time
h=half life
A=final amount
so

A=0.242188 grams

Answer 2

Answer:

0.24 grams of polonium is in the sample 966 days later.

Step-by-step explanation:

Given : Suppose a laboratory has a 31 g sample of polonium-210. The half-life of polonium-210 is about 138 days.

To find : How many half-lives of polonium-210 occur in 966 days? How much polonium is in the sample 966 days later?

Solution :

The half-life of polonium-210 is about 138 days.

We have to find the number of half-lives of polonium-210 occur in 966 days.

The number of half-lives is $n=\frac{966}{138}=7$

The amount of polonium-210 remaining after t days is given by the equation,

$N(t)=N_0\times (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}$

where,

N(t)  is the amount of substance remaining after  t  days,

$N_0=31$ is the initial amount of substance,

t=966 is the time in days,

$t_{\frac{1}{2}}=138$ is the half-life of the substance.

Substitute the values in the formula,

$N(t)=31 \times (\frac{1}{2})^{\frac{966}{138}$

$N(t)=31 \times (\frac{1}{2})^{7}$

$N(t)=31 \times 0.0078125$

$N(t)=0.24$

Therefore, 0.24 grams of polonium is in the sample 966 days later.